similar to: smbclient -L succeeded even network is down

Hello list, I'm running a samba 3.5.3 CTDB cluster, found the output is different Q1: What does the "0:" mean in pid column? There was no such stuff in non-CTDB smbstatus output. <snip> samba_01:~ # smbstatus -S 2>/dev/null Service pid machine Connected at ------------------------------------------------------- ben 0:21363 samba Mon Oct

The following dummy data frame has factor Q (with 2 levels) nesting factor P (with levels p1 and p2 nested under q1, and p3 and p4 nested under q2), but both crossing the random variate s, which has 8 levels. The dependent measure is dv. > # The data frame: > testnest dv s P Q 1 1 s1 p1 q1 2 2 s2 p1 q1 3 1 s3 p1 q1 4 2 s4 p1 q1 5 1 s5 p1 q1 6 3 s6 p1 q1 7 3 s7

Dear R-help, I wonder if you could give me some suggestions in how to do a union join of two data frames as follow: -> union join the common column, and insert a 0 if one is missing. I made a function to perform the following, and I know it may not that quite welly written, but it works. Any suggestions are welcome, many thanks. Anthony > q1 =

----- Original Message ----- From: "Silvano" <silvano at uel.br> To: <r-help at r-project.org> Sent: Thursday, June 30, 2011 9:07 AM Subject: Tables and merge > Hi, > > I have 21 files which is common variable CODE. > Each file refers to a question. > > I would like to join the 21 files into one, to construct > tables for each question by CODE. >

sample report data that i want to forecast quarter quarter_index Revenue 2007 Q1 1 $3,856,799 2007 Q2 2 $4,243,328 2007 Q3 3 $4,930,369 2007 Q4 4 $5,443,579 2008 Q1 5 $5,164,830 2008 Q2 6 $5,104,413 2008 Q3 7

sample report data that i want to forecast quarter quarter_index Revenue 2007 Q1 1 $3,856,799 2007 Q2 2 $4,243,328 2007 Q3 3 $4,930,369 2007 Q4 4 $5,443,579 2008 Q1 5 $5,164,830 2008 Q2 6 $5,104,413 2008 Q3 7

I have a data.frame named "df". The dput of df is at the bottom of this e-mail. What I'd like to do is replace the "n/a " values with NA. On Mac OSX, it works to do this: df[df == "n/a"] <- NA However, it does not work on Ubuntu. See below. Thanks in advance, Garrett > x <- df[27, 4] # complete data.frame dput is below > dput(x) "n/a?"

Hi Guys, I'm using zoo package now. I found lag is not doing what I assumed. > x <- zoo(11:21) > z <- zoo(1:10, yearqtr(seq(1959.25, 1961.5, by = 0.25)), frequency = 4) > x 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 15 16 17 18 19 20 21 > lag(x) 1 2 3 4 5 6 7 8 9 10 12 13 14 15 16 17 18 19 20 21 > z 1959 Q2 1959 Q3 1959 Q4 1960 Q1 1960 Q2 1960 Q3 1960 Q4

Example: Will ts objects be obsolete or modified? > a [,1] 1983 Q1 2.747365190 1983 Q2 2.791594762 1983 Q3 -0.009953715 1983 Q4 -0.015059485 1984 Q1 -1.190061246 1984 Q2 -0.553031799 1984 Q3 0.686874720 1984 Q4 0.953911035> lag(a,4) [,1] 1983 Q1 NA 1983 Q2 NA 1983 Q3 NA 1983 Q4 NA 1984 Q1 2.747365190 1984 Q2

I have a xts object with logical data . > tail(q1) ..1 2010-02-19 TRUE 2010-02-22 FALSE 2010-02-23 FALSE 2010-02-24 FALSE 2010-02-25 FALSE 2010-02-26 FALSE > I want to build a xts that records the dates that there is a change. If TRUE -> FALSE it is "down" if FALSE -> TRUE it is "up" I can do it like this... q2 <- matrix(ncol=2, nrow =

Hi, Why the qr() produces a negative Q compared with Gram-Schmidt? (note example below, except Q[2,3]) Here is an example, I calculate the Q by Gram-Schmidt process and compare the output with qr.Q() a <- c(1,0,1) b <- c(1,0,0) c <- c(2,1,0) x <- matrix(c(a,b,c),3,3) ########################## # Gram-Schmidt ########################## A <- matrix(a,3,1) q1 <-

I have the initial matrice: > *data.frame(Subject=rep(100:101, each=4), Quarter=rep(paste("Q",1:4, sep=""),2), Boolean = rep(c("Y","N"),4))* Subject Quarter Boolean 1 100 Q1 Y 2 100 Q2 N 3 100 Q3 Y 4 100 Q4 N 5 101 Q1 Y 6 101 Q2 N 7 101 Q3 Y 8 101

Greetings, I'm learning R and I'm stuck on a basic concept: how to specify a logical condition once and then perform multiple transformations under that condition. The program below is simplified to demonstrate the goal. Its results are exactly what I want, but I would like to check the logical state of gender only once and create both (or any number of) scores at once.

Hello RoR! I''m building a rails app for hosting multi-page surveys. I set up questionnaire, question, and question_component models. Now I''m initializing them with some data. survey.rb q1 = Question.new "q1" q1.quote "Welcome to a test of the new interviewing system." q2 = Question.new "q2" q2.quote "Do you like pie?"

I am generating an ftable (by running ftable on the results of a xtabs command) and I am getting the following. ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?Var1? Var2 date ? ? ? ? ? ? ? ? group? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2007-01-01? ? ? ? ? q1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?1? ? 9 ? ? ? ? ? ? ? ? ? ? ?q2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

Hi, I have a question about how to obtain the union of several data frame rows. I'm trying to create a common key for several tests composed of different items. Here is a small scale version of the problem. These are keys for 4 different tests, not all mutually exclusive: id q1 q2 q3 q4 q5 q6 1 A C 2 B D 3 A D B 4 C D B D I would like

Hello, (Apologies for length, wanted to get all the relevant detail in that I know of). I've been having a lot of trouble with some code for an inventory analysis problem I was doing, and finally came to the conclusion that it appears that choose() is returning incorrect values. Specifically: ------------- Browse[1]> nn [1] 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3

I have the initial matrice: > *data.frame(Subject=rep(100:101, each=4), Quarter=rep(paste("Q",1:4, sep=""),2), Boolean = rep(c("Y","N"),4))* Subject Quarter Boolean 1 100 Q1 Y 2 100 Q2 N 3 100 Q3 Y 4 100 Q4 N 5 101 Q1 Y 6 101 Q2 N 7 101 Q3 Y 8 101

Hello, I have a question that is not directly related to R ... but I try to do it in R ;-) : I would like to generate a matrix Q satisfying (for a given Z, X and W) the two following conditions: t(Q)%*%Q=Z (1) XQ=W (2) where: Q is m rows and r columns X is p rows and m columns D is p rows and r columns C is r rows and r columns with m>p,r e.g: m=6, p=2 r=3

hi, for example, I have an string "q1 q2 q3". how can I encode it in rails to produce the result like "q1%20q2%20q3". Thanks! Best Regards, Zhenjian -------------- next part -------------- An HTML attachment was scrubbed... URL: http://wrath.rubyonrails.org/pipermail/rails/attachments/20060717/3824f992/attachment.html